package algorithm.problems.string;

import java.util.StringTokenizer;

/**
 * Created by pradhang on 7/11/2017.
 * Compare two version numbers version1 and version2.
 * If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.
 * <p>
 * You may assume that the version strings are non-empty and contain only digits and the . character.
 * The . character does not represent a decimal point and is used to separate number sequences.
 * For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.
 * <p>
 * Here is an example of version numbers ordering:
 * <p>
 * 0.1 < 1.1 < 1.2 < 13.37
 */
public class CompareVersionNumbers {
    /**
     * Main method
     *
     * @param args
     * @throws Exception
     */
    public static void main(String[] args) throws Exception {
        System.out.println(new CompareVersionNumbers().compareVersion("1.11.1", "1.11"));
    }

    public int compareVersion(String version1, String version2) {
        StringTokenizer st1 = new StringTokenizer(version1, ".");
        StringTokenizer st2 = new StringTokenizer(version2, ".");
        while (st1.hasMoreTokens() & st2.hasMoreTokens()) {
            int token1 = Integer.parseInt(st1.nextToken());
            int token2 = Integer.parseInt(st2.nextToken());
            if (token1 > token2)
                return 1;
            else if (token2 > token1)
                return -1;
        }
        if (st1.countTokens() > st2.countTokens()) {
            while (st1.hasMoreTokens()) {
                if (Integer.parseInt(st1.nextToken()) > 0)
                    return 1;
            }
        } else if (st2.countTokens() > st1.countTokens()) {
            while (st2.hasMoreTokens()) {
                if (Integer.parseInt(st2.nextToken()) > 0)
                    return -1;
            }
        }
        return 0;
    }

}
